It is known that a closed (ie. connected, compact without boundary) surface is homeomorphic to a sphere or to a connected sum of finitely-many tori or projective spaces. Let $S_g$ (resp. $\Sigma_g$) denote the connected sum of $g$ tori (resp. projective spaces). Because $\chi(A \sharp B)= \chi(A)+ \chi(B)-2$, we easily deduce the Euler characteristics $\chi(S_g)=2-2g$ and $\chi(\Sigma_g)=2-g$; the integer $g$ is called the genus. A surface group is a group isomorphic to the fundamental group of one of these surfaces. If $G$ is a surface group, two cases happens: either $G$ is the fundamental group of an orientable surface $S_g$ of genus $g$ and

$G = \langle a_1,b_1, \dots, a_g,b_g \mid [a_1,b_1] \cdots [a_g,b_g]=1 \rangle$,

or $G$ is the fundamental group of a non-orientable surface $\Sigma_g$ of genus $g$ and

$G = \langle a_1, \dots, a_g \mid a_1^2 \cdots a_g^2=1 \rangle$.

(Just write $S_g$ or $\Sigma_g$ as a polygon with pairwise identified edges, and apply van Kampen’s theorem; for more information, see Massey’s book Algebraic Topology.)

Our main result is the following characterization of subgroups of a surface group:

Theorem 1: Let $G$ be the fundamental group of a closed surface $S$ and $H$ be a subgroup. Either $H$ has finite index $h$  in $G$ and $H$ is isomorphic to the fundamental group of a closed surface whose Euler characteristic is $h \cdot \chi (S)$, or $H$ is an infinite-index subgroup of $G$ and is free.

For this note, I was inspired by Jaco’s article, On certain subgroups of the fundamental group of a closed surface. The proof of property 2 below comes from Stillwell’s book, Classical topology and combinatorial group theory.

Essentially, the theorem above follows from Property 2 below:

Property 2: The fundamental group of a non-compact surface is free.

Combined with Property 3, let us first notice some corollaries of Theorem 1 on the structure of surface groups.

Property 3: The abelianization of $\pi_1(S_g)$ (resp. $\pi_1(\Sigma_g)$) is isomorphic to $\mathbb{Z}^{2g}$ (resp. $\mathbb{Z}^{g-1} \times \mathbb{Z}_2$). Therefore, two closed surfaces $S_1$ and $S_2$ are homeomorphic if and only if their fundamental groups are isomorphic.

Proof. In order to compute the abelianizations, it is sufficient to add all the possible commutators as relations in the presentations given above. Therefore,

$\pi_1(S_g)^{ab} \simeq \langle a_1,b_1 , \dots, a_g, b_g \mid [a_i,a_j] = [b_i,b_j ] = [a_i,b_i]=1 \rangle \simeq \mathbb{Z}^{2g}$

and, with $z: = a_1\cdots a_g$,

$\pi_1( \Sigma_g)^{ab} \simeq \langle a_1, \dots, a_{g-1},z \mid [a_i,a_j] = [a_i,z]=1, \ z^2=1 \rangle \simeq \mathbb{Z}^{g-1} \times \mathbb{Z}_2$.

Consequently, the $\pi_1(S_g)$ and $\pi_1( \Sigma_1)$ define a family of pairwise non-isomorphic groups, and the conclusion follows from the classication of closed surfaces. $\square$

Corollary 1: Let $G$ be the fundamental group of a closed surface different from the projective space. Then $G$ is torsion-free.

Proof. The abelianization of a surface group is cyclic if and only if it is the fundamental group of the projective plane; therefore, the same conclusion holds for the surface groups themself, and we deduce that $G$ has no cyclic subgroup of finite index. Consequently, the cyclic subgroup generated by a non trivial element of $G$ is free, that its order is infinite. $\square$

Corollary 2: Let $G$ be the fundamental group of a closed surface $S$ satisfying $\chi(S) \leq 0$ and $H$ be a subgroup generated by $k$ elements. If $k < 2- \chi (S)$ then $H$ is free.

Proof. If $\chi(S)=0$, $S$ is a torus or a Klein bottle, and the statement just says that its fundamental group is torsion-free, that follows from Corollary 1.

From now on, we suppose that $\chi(S)<0$. According to Property 3, the abelianization of $G$ has rank $2- \chi(S)$. We deduce that the smallest cardinality of a generating set of $G$ has cardinality $2- \chi(S)$; in particular, $H \subsetneq G$ because $k < 2- \chi(S)$. Suppose that $H$ has finite index $h$. Then $H$ is the fundamental group of a closed surface $\Sigma$ and $\chi(\Sigma)= h \cdot \chi(S)$. In the same way, necessarily $k \geq 2- \chi(\Sigma)$. Therefore,

$k \geq 2- \chi(\Sigma) = 2-h \cdot \chi(S) > 2 - \chi(S)$,

a contradiction. $\square$

Corollary 3: Let $G$ be the fundamental group of a closed surface $S$ satisfying $\chi(S)<0$. If $x,y \in G \backslash \{1\}$ commute, then there exist $z \in G$ and $m, n \in \mathbb{Z}$ such that $x=z^n$ and $y=z^m$.

Proof. Let $g \geq 2$. Then it is easy to find two epimorphisms

$\pi_1 (S_g ) \twoheadrightarrow \langle a_1 , b_1 , a_1 , b_2 \mid [ a_1, b_1 ] = [ a_2, b_2 ] ^{-1} \rangle \simeq \mathbb{F}_2 \underset{\mathbb{Z}}{\ast} \mathbb{F}_2$

and

$\pi_1 ( \Sigma_g ) \twoheadrightarrow \langle a_1, a_2 \mid a_1^2 = a_2^{-1} = 1 \rangle \simeq \mathbb{Z}_2 \ast \mathbb{Z}_2$.

Therefore, the groups $\pi_1(S_g)$ and $\pi_1(\Sigma_g)$ are not abelian for $g \geq 2$, and we deduce that the sphere, the projective plane and the torus are the only closed surfaces whose fundamental group is abelian.

Because $S$ satisfies $\chi(S)<0$, we conclude that $G$ has no finite-index abelian subgroup, and that the subgroup genereted by $\{ x,y \}$ is necessarily free; since $x$ and $y$ commute, the subgroup turns out to be cyclic and the conclusion follows. $\square$

Corollary 4: The commutator subgroup of a surface group is free.

Proof. Let $G$ be the fundamental group of a closed surface $S$. If $S$ is a projective space, then $G \simeq \mathbb{Z}_2$ and its commutator subgroup is trivial (in particular free). Otherwise, thanks to Property 3 we know that the abelianization of $G$ is infinite; therefore, the commutator subgroup is an infinite-index subgroup and so is free. $\square$

Proof of property 2. First, we notice that the fundamental group of a connected compact surface with boundary is free. If $S$ is such a surface, by gluing a disk along each boundary component, we get a closed surface $\overline{S}$; therefore, $S$ is homotopic to a punctured closed surface. Because a closed surface may be identified with a polygon whose edges are pairwise identified, it is not difficult to prove that a punctured closed surface is homotopic to a graph (by “enlarging the holes”); in particular, we deduce that $\pi_1(S)$ is free.

The figure below shows that the fundamental group a 2-punctured torus is a free group of rank three. From now on, let $S$ be a non-compact surface. In order to prove that $\pi_1(S)$ is free, we want to find a sequence of compact surfaces with boundary

$S_1 \subset S_2 \subset \cdots \subset S$

such that $S= \bigcup\limits_{i \geq 1} S_n$ and  that the inclusions $S_i \hookrightarrow S_{i+1}$ are $\pi_1$-injective. It is sufficient to conclude since we proved above that the fundamental group of a compact surface with boundary is free.

Take a triangulation of $S$ so that $S$ may be identified with a simplicial complex; let $\Delta_1,\Delta_2, \dots$ denote the 2-simplexes. Without loss of generality, we may suppose that $\Delta_{i+1}$ is adjacent to one of the simplexes $\Delta_1, \dots , \Delta_i$; otherwise, number the simplexes by taking first the simplexes adjacent to a vertex $P$ cyclically, then the simplexes within $1$ from $P$, then the simplexes within $2$ from $P$, etc.

Notice that a connected union of simplexes may not be a surface; however, if $\Delta_i'$ denotes the union of $\Delta_i$ with a small closed ball around each of its vertices, a connected union of $\Delta_i'$ is automatically a compact surface with boundary. Now, we construct the surfaces $S_n$ by induction:

Let $S_1=\Delta_1'$. Now suppose that $S_n$ is given. If $\Delta_{n+1}' \subset S_n$, then set $S_{n+1}=S_n$; otherwise, define $S_{n+1}$ as the union of $S_n$ with $\Delta_{n+1}'$ and with the disks bounding a boundary component of $S_n \cup \Delta_{n+1}'$.

Because $S_n$ is a union of $\Delta_i'$, we know that it is a compact surface with boundary. To conclude, it is sufficent to prove that the inclusion $S_i \hookrightarrow S_{i+1}$ is $\pi_1$-injective. To do that, just notice that $\Delta_{n+1}'$ is attatched on the boundary of $S_n$ in six possible ways: In the first three cases, $S_{n+1}= S_n \cup \Delta_{n+1}'$ because the boundary component bounds a disk in $S$ if and only if it bounds a disk in $S_n$. So $S_{n+1}$ clearly retracts on $S_n$ so that the inclusion $S_n \hookrightarrow S_{n+1}$ induces an isomorphism $\pi_1(S_n) \simeq \pi_1(S_{n+1})$.

In the three last cases, up to homotopy, we just glue one or two edges on a boundary component, and using van Kampen’s theorem, we notice that a free basis of $\pi_1(S_{n+1})$ is obtained by adding one or two elements to a free basis of $\pi_1(S_n)$. $\square$

Proof of theorem 1. Let $\overline{S} \to S$ be the covering associated to the subgroup $H$; in particular, $\overline{S}$ is a surface of fundamental group $H$. If $H$ is a finite-index subgroup, $\overline{S}$ is compact and the conclusion follows. Otherwise, $\overline{S}$ is not compact, and the conclusion follows from Property 2. $\square$

In the note Some SQ-universal groups, we prove also that almost all surface groups are SQ-universal. In particular, it implies that they have uncountably many norma subgroups.